CMAT Number Systems Questions

It is given that $$\frac{1}{x}$$ is positive and $$\frac{1}{y}$$ is negative, this implies that x is positive and y is negative.

A) $$\frac{1}{x}\ +\ \frac{1}{y}$$ is positive, we know x is positive, and y is negative, but we cannot conclude which value is greater. Therefore, this may or may not be true.

B)$$\frac{1}{x}-\frac{1}{y}$$ is positive; we know that x is positive and y is negative. So this statement is definitely true.

C) $$\ \frac{\ x-y}{xy}$$= -($$\frac{1}{x}-\frac{1}{y}$$)

As $$\frac{1}{x}-\frac{1}{y}$$ is positive, $$\ \frac{\ x-y}{xy}$$ is negative and statement is true.

D) $$\frac{1}{xy}$$ is negative as x is positive and y is negative. Therefore, the statement is false.

E)$$\frac{1}{x^2}-\frac{1}{y^2}$$ is positive only when $$y^2>x^2$$, but we cannot conclude which value is greater. Therefore, this may or may not be true.

Only (B) and (C) are true.

Answer is option B

We have dimensions : (24*18)

HCF ( 24, 18) = 6

Now Area of floor: 432

In order to minimise the number of tiles considering a tile with a dimension equivalent to the HCF,  we have the dimension of the tile to 6*6 = 36 units.

A total of $$\frac{432}{36}$$  = 12 tiles are required.

Minimum number of tiles that pave the entire floor :(LCM)^2/Area = 12

We have :
$$-\frac{4}{5}\ ,\ -\frac{5}{12}\ ,\ -\frac{7}{18}\ ,\ -\frac{2}{3}$$
Now LCM is 180
so we get :
$$-\frac{144}{180}\ ,\ -\frac{75}{180}\ ,\ -\frac{70}{180}\ ,\ -\frac{120}{180}$$
Now arranging in ascending order
we get :
$$-\frac{144}{180}\ <\ -\frac{120}{180}<-\frac{75}{180}<-\frac{70}{180}$$
so order is A<D<B<C

According to question,

Number AXYZ/9 = XYZ

It can be written as

A000/9 + XYZ/9 = XYZ

A000/9 + XYZ/9 = 9XYZ/9

A000/9 = 8XYZ/9

A000 = 8XYZ

We have to find the maximum value of A so that when A000 is divided by 8 gives three digit numbers.

7000/8= 875 and 8000/8=1000.

So, answer is 7.

Prime factorization of :

$$2400=(2)^5\times(3)\times(5)^2$$

Now, to make it a perfect cube, we need to make the powers multiple of 3, and thus we need to divide it by = $$(2)^2\times3\times(5)^2=300$$

=> Ans - (D)

Let the two numbers be $$9x$$ and $$9y$$, where $$x$$ and $$y$$ are co prime numbers.

=> $$9x+9y=135$$

=> $$x+y=\frac{135}{9}=15$$

Now, pairs of natural numbers having sum 15 and co prime are = (1,14), (2,13), (4,11), (7,8)

Thus, 4 such pairs can be formed.

=> Ans - (C)

According to BODMAS,

first Multiply the given numbers

0.3 x0.3= 0.09

0.03 x0.03=0.0009

0.6 x 0.03=0.018

simplifying this according to the given question,

0.09-0.0009-0.018=0.0729

now $$\ \frac{\ 0.0729}{0.54}$$

=> 0.135

Therefore answer is option b 

Let the two numbers be $$x$$ and $$4x$$

=> $$x\times4x=84\times21$$

=> $$x^2=(21)^2$$

=> $$x=21$$

$$\therefore$$ Larger number = $$4\times21=84$$

=> Ans - (D)

Let the number be 'x'

one-fourth of the number = $$\frac{1}{4}\times x = \frac{x}{4}$$

one-third of one-fourth of the number = $$\frac{1}{3}\times \frac{x}{4} = \frac{x}{12}$$

Given, one-third of one-fourth of the number is = 15

$$\Rightarrow \frac{x}{12} = 15$$

$$\Rightarrow x = 180$$

Hence the number is 180.

Then three-tenth of the number = $$\frac{3}{10}\times x = \frac{3}{10}\times 180 = 54$$

The topper of the class is the one who gets a highest overall score. Let us check the options, who has got the overall score.

Option A:

The overall score of Maya = 90 + 50 + 90 + 60 + 70 + 80 = 440 marks 

Option B:

The overall score of Arya = 100 + 80 + 88 + 60 + 80 + 70 = 478 marks

Option C:

The overall score of Rupesh = 80 + 55 + 85 + 95 + 50 + 90 = 455 marks

Option D:

The overall score of Riya = 90 + 72 + 70 + 72 + 91 + 70 = 465 marks

So, the topper of the class is Arya, with a overall score of 478 marks.

$$360\sqrt?$$ + 4500 = 19700 - 9800

$$\Rightarrow 360\sqrt? + 4500 = 9900$$

$$\Rightarrow 360\sqrt? = 5400$$

$$\Rightarrow \sqrt? = 15$$

$$\Rightarrow ? = 15^2$$

$$\Rightarrow ? = 225$$

$$\sqrt{188+ \sqrt{51+ \sqrt{169}}}$$ = $$\sqrt{188+ \sqrt{51+13}}$$ = $$\sqrt{188+ \sqrt{64}}$$ = $$\sqrt{188+ 8}$$ = $$\sqrt{196}$$ = 14